Answer
$f^{-1}(x)=\frac{-3-x}{x-2}$
Work Step by Step
$f(x)=\frac{2x-3}{x+1}$
$y=\frac{2x-3}{x+1}$
a.
$x=\frac{2y-3}{y+1}$
$xy+x=2y-3$
$xy-2y=-3-x$
$y(x-2)=-3-x$
$y=\frac{-3-x}{x-2}$
$f^{-1}(x)=\frac{-3-x}{x-2}$
b.
$f(f^{-1}(x))=\frac{2\left(\frac{-3-x}{x-2}\right)-3}{\frac{-3-x}{x-2}+1}=\frac{-5x}{-5}=x$
and
$f^{-1}(f(x))=\frac{-3-\frac{2x-3}{x+1}}{\frac{2x-3}{x+1}-2}=\frac{-5x}{-5}=x$
